1
-work
(3.3.4)
①
cylinder
pistone
Frictionless
system
1kg
P1
"
steam
Assumptions:
Isobaric
CPE
3). HE
No
4).
questions:
=
125°
2 V,
=
P1 P2
=
->
be neglected
can
shaft
101.3 kPa 0.1 MPa
=
=
V2
Closed system
11.
bar
1.013
=
T1
27.
(constant pressure)
work
of steam?
A?
II?
Tz?
H?
work
Answer:
Mass Balance:AM= M2-M1
Energy
0
=
M2 M1 M
<=>
=
=
(U+M(=2 w
Balance:
a
+
McLK
-
M, L,
G-POV
MI ) G -JPa q
M(1-1i) 9-PM/vi vi)
=
=
=
-
From
P 0.1 MPa
p)ar q
-
PM
+
-
=
-
steam table
#
Y
T,°
I
100
1.6950
2506.7
2676.2
150
1.9364
2582.0
2776.4
=
4100 I
is
-
=
150
-
100
I
!095050-500
0.10)
=
-
p(v x)
-
M((ui+pvi) -(+p)]
(v2 vi)
M(H2 H)
:
=
-
9 M(( ()
9
par
=
-
=
+
Vi
=
1.1 i
=
i
/member
I-
-%
For
1
12506.7-=15
2582.8
(125
We
can
-
2506.7
=
2776.4
=
calculate:I
2M 2(7.8161)
=
=
We know P2 0.1 MPa,
=
so
we can
I
T,
600
4.028
3301.43704.7
For
virus soe
-5463
T2
-
steam table
#
3131.6
72
from
3488.1
3.565
=
3.6322
=
lookfor to
500
6.72
2676.2
e
2506.7 2544.75
+
=
-
-1500 (5-8e
=
34o8
12
500
e
3408.1 + 31.407
=
12 3519.507
=
514.5°,
=
11
ore
1,500 25
-
-33
112
a
M(H
=
e
24.6935
=
-
+]
3131.6
+
3156.244/
=
yy[3519.507 2726.37 743.207k]/
=
-
=
Work
W
of
=
-
=
-
Steam
(PaV
=
-
p(ic v)
-
1.8161)
7.013x10549(3.6322-
=-
x105]/kg
1839
-183.
=
=
kJ/ky,
Energy Changes
Al (72 4i 3156.294-2544.75 611.54443/kg
Internal
=
-
=
=
Enthalpy Changes
Ati H2
=
Steam
13.3.1)
+1, 3514.507-2726.3
793.207
=
-
=
K)/ ky
ideal gas
as
Assumption:constant
(P1=P2)
pressure
ideal gas:PX=
NRT
heat
c>
=
N
capacity(p 34.4)/mol/K
=
=
112 2 XI
=
T1 125 273.15
Ni N2
=
=
*
2
+
T2 2
=
*T T12
Mole
IELT
N
=
9 NAH
=
7608327
-(PV
=-
=
of
Ikg
the
Steam
I gYmol=
=
55.55
=
mol
NCp AT
=
=
(390.15) 796.3K,
=
=55.55 mol
X
390.15k
=
=
55.55
"1796.3-390.15)K
34.4
x
760.032k]/
-
PAX
molx
=-1838027
-
=
P(Xz x)
-
0.314
=
m
p P, P2
Amber:
=
=
-
=
4(* *)
-
(796.3-340. 151K
-183.082
k],
=-NR
(T2 -T)
=
②
(3.3.b)
Frictionless
"
P,
Closedsystem
KE & PE
3).
Volume
0
=
be neglected
constant
system
work
-Mi
m
=
=
can
shaft
No
1 (v ME*4:2):9/wip
Balance:
Energy
+
Mc=
M,i,
-
P2 241 2(101 bar)
=
vi
P2 2 PI
=
2).
M
volume constant
steam
1.013 bar
=
T1 125 °
4).
Mass Balance:
kg
known:1
system
Assumptions:1).
Cylinder
Pistone
=
known
=
bar
2.026
=
V,
from (3.3.4)
-()/
a
=
1.8761
=
-
0.2026
=
MPa
Vi (constant volume)
=
From Steum Table:
Y
i
T,C
1.7014
500
2.013
600
3130.8
3301.4
I
by interpolation
(P
0.1490
=
514.98°,
from
Is.3. a)
9 1kg (3156.36
=
-
1
that
2544.75
=
2544.75)E
a
Y
(170.6)
+3130.8
3156.3643/k9x
14.98
+
=
known
getis
=
=
500
=
can
-3130,85 40-8et
sogotothe
12
From here, we
k3/Ag
611.61K]/
=
Steam
13.3.1)
as
ideal gas
N1
X Vz
=
42
2
=
a
=
N2
=
=
4
(p 34.47/mol/k
the volume constant.
known:T1
*2
125° 390. 15K
=
=
=
i
From
2
=
=
=
Balance:G NAL NC-AT
Energy
a
/Member
2(398.15) 796.3k
T
=
=
N(<p-R) (Tc -T1)
=
mx (34.4
55.55
+
=
-
8.314(m"(796.3-390.15)
5769507 576.95k]/
=
=
③
Thomson
13.5) Joule
Expansion
KNOWn:
is
Assumptions:11.
Steam
flows
rapidly
PEC KE
can be
3).
Flow is
1).
No
61.
Energy
1
P2
shaft
=
1
=
bar
0.1 MPa
=
questions:
0
=
neglected.
T?
is
volume
0 10
=
0
+
50 MPa
=
T=600"
steady
4). Constant
Energy
i
-
2).
Mass Balance:
41:500 bar
Steam
if
the gas is
work
balance
0
=
M2
=
-M1
<>
=
t(IM**Y]:
Balance:
2M
0
Mc M1 M
=
(fn'+w,"w
MH M11
=
=
M(Hi- +)
=
AA
=
-
Mctz
ideal?
Initial&
final
States Steam have the same
enthalpy.
H(T, 41) H(T2, P2)
=
We
can
from steam table,
is
find
I
T,°C
50 MPa
3247.6
600
[3908
0.1MPa
by interpolation
look for
->
with
this value in 0.1 MPa
3074.3
3270.2
307e
g -638
12-300
12
Ifthe
(
)
100+
=
gas
04.99
=
385°/
304.990)
=
=
is ideal.
Joule Thomson coefficient
for ideal gas is zero,
so
the
enthalpy
of the gas is only
dependenton temperature.
H (T) H(Tz)
=
T1
T2
=
600°C,
=
Water to be heated
④ (3.6).
in
an
open metal drum.
i system
Steam,
Mwater 100
=
1 i
=
T2 80"
=
Pin= 3 bar=0.3 MPa
Tin= 300"
kg MT
=
Assumptions:11.
& lose is
neglected
->
0
shaftwork
2).
No
3).
Volume constant
4).
PEC KEare
neglected.
0
=
Question:
Msteam
(AM)?
Mass Balance:AM Miz
=
MT,
-
it[U+METY
Balance:
Energy
(M)
From
steam
IV/T,
0.001
Hin
003
0.001024
=
K3
3069.3
=
(ii) +,
A)
104.80
=
(ii) T2
(MI) is
Mizlin
1.003
=
mg
x
10
m"/kg
1,024 x10-3
/ky
I
/Ag
Note: Internal energy of Liquid does not depend
( M+2 M+) Hin
(MI) +,
-
=
-
MinFin M+,LT,
-
=
-
((Yin Hin)
-
M M+,
=
=100
MT,
=
Mt,Fin
(LT Hin)
-
)
100
=
1-2964.42)
x
Miz
=
x
100.4)kyx
=
-
=8.41
My
kg
108.
=
steam
3069.3)
1334.86
1-2734.44)
AM
+
1 MI) +=AMAin
=
k3/Ag
334.86
=
Miz
(HYY in
table:
=
(V) +2
-
[Mx
=
41-100
added
/
-
3069.3)
on
pressure
③ (3.16)
Adiabatic Turbine
Steam
T2 200°
=
7=500°
P2 0.3 MPa
=
P, 3.5 MPG
=
Hi
+=3450.
WS: -750kW
of
Assumptions:11. Steady
(4). Mass
state
Volume system constant
3).
Adiabatic
4).
PERKE
is
Balance:
1
0 E)
=
->
0
0
Balance:
-
open
system
0
=
neglected.
M1 M2
+
=
M2
Energy
-750x1033/s
Steam?
question:Flowrate
2).
=
2865.6
=
=
Mi
-
[M/FtiTh+iwis -
t (* **5]:
0
Wi
wi
Mit Mic +iz W's
MiHi
MCH2
=
-
=
=
-
-
M,HY Mit
+
M(Hi Hi)
Wi
M
+
+
=
-
W) o.4)
=
=
583
-
M1
1.2014 Ky
=
/s
//
(b).
If
there is
T
a
=$G
heat loss
=
-60k1
/Ag
500°
T2 150°
=
=
P2 0.3 MPa
=
P, 3.5 MPC
=
12
11 3450. k)/Ag
=
9
-
=
C
Energy
Balance:
0 =0
=
t
M1+ M2
=
=
k]/Ag
60k3/ky
=?
Ws
Mass Balance:
276)
=
M1
=
M2
(Ying 2M (fg +g+ws-pe
=
0
Ns
M, H,
=
=
McHc 0 Ws
+
+
+
-MiH -M2H2 -G
Mi (H2- Hi)
=
-
+
J
=
=-
=
-
084.03
807
-
kW/
-
0
3450.9) 1 60 1.281449)
k1/+ 76.084k]/S
807.15k]/S
-
+
4
1.2814kg(2761
=
-MiH1 MiHz
=
=
-
x